Doubts in finding the range of a function

Question

I am solving function $\sqrt{4-x}$ to find the range.

I have learnt two methods to solve online with few doubts.

First method -

Let function y = $\sqrt{4-x}$

Then,

$y^2 = 4-x$

$x = 4-y^2$

Now can we find range $4-y^2 \ge 0$ in this way?

$y \le \pm 2$

Then what is the range? Is above equation correct?

Second method says range is equal to domain of inverse of a function.

so it says to interchange y with x,

then we have,

$x = \sqrt{4-y}$

Then solve for y,

$x^2 = 4-y$

$y = 4-x^2$

then how to find range further?

Answer 1

The range of $f$ is the set of those numbers $y$ such that the equation $f(x)=y$ as some solution. But\begin{align}f(x)=y\iff&\sqrt{4-x}=y\\\iff&4-x=y^2\text{ and }y\geqslant0\\\iff&x=4-y^2\text{ and }y\geqslant0\end{align}and therefore the range of $f$ is $[0,+\infty)$.

Answer 2

Instead of looking for some hard and fast way to find the domain and range, try and think it out.

If $x > 4$, you get a negative under the square-root. Hm. That's not good! So I mustn't be allowed to let $x$ be bigger than $4$. What if $x \leq 4$? Well, I sure think it's defined for all everything less than $4$. Is $4 - x$ always non-negative for $x \leq 4$. You bet it is. So, the domain must be $(-\infty, 4]$.

Now, I know that as we make $x$ smaller and smaller (starting just before $x = 4$) my value of the function gets bigger and bigger. What happens to our function at our upper limit of $x$, $ x = 4$? Well, it's $0$. So, the range must be $[0, \infty)$ as I sweep left and right across my allowable domain.

The range of the function depends completely on the domain of the function. After you reason out the domain of the function ask yourself,

1. What is the value of my function at the lowest value of my domain.
2. What is the value of my function at the biggest value of my domain.
3. Are there any values in my domain which make the function discontinuous? That is, does there a value (or even values) for $x$ that create a 'hole(s)' in my function when I plot it?

Answer 3

Let $f(x) = \sqrt{4 - x}$. Since we require that $4 - x \geq 0 \implies 4 \geq x$, the implied domain of $f$ is $\text{Dom}_f = (-\infty, 4]$.

Since $\sqrt{4 - x}$ denotes the principal, or nonnegative, square root of $4 - x$, $\sqrt{4 - x} \geq 0$. Notice that as $x$ becomes a large negative number, $4 - x$ becomes a large positive number, so $\sqrt{4 - x}$ also becomes a large positive number. As $x \to -\infty$, $4 - x \to \infty$, so $\sqrt{4 - x} \to \infty$. Thus, the range of $f$ is $\text{Ran}_f = [0, \infty)$.

Let's consider your approach. As stated above, $y = \sqrt{4 - x} \implies y \geq 0$. Solving for $x$ yields \begin{align*} y & = \sqrt{4 - x}\\ y^2 & = 4 - x\\ x & = 4 - y^2 \end{align*} The graph of $x = 4 - y^2$ is a parabola with vertex $(4, 0)$ that opens to the left. However, the restriction that $y \geq 0$ means that the graph only consists of the upper half of the parabola, as shown below.

From the graph, we see that the domain of $f$ is $\text{Dom}_f = (-\infty, 4]$ and the range of $f$ is $\text{Ran}_f = [0, \infty)$ since $y$ will take on increasingly large values as $x$ takes on increasingly large negative values.

As for your idea of determining the domain by finding the inverse, keep in mind that the domain of the inverse function is the range of the function.

Let's solve for the inverse. Again, we need to keep in mind that $y = \sqrt{4 - x} \implies y \geq 0$. \begin{align*} f(x) & = \sqrt{4 - x}\\ y & = \sqrt{4 - x}\\ y^2 & = 4 - x\\ x & = 4 - y^2, y \geq 0 \end{align*} Interchanging variables yields $$y = 4 - x^2, x \geq 0$$ so our inverse function is $$g(x) = 4 - x^2, x \geq 0$$ which has domain $\text{Dom}_g = [0, \infty)$ and range $\text{Ran}_g = (-\infty, 4]$. Since the domain of the inverse function is the range of the function and the range of the inverse function is the domain of the function, we conclude that the domain of $f$ is $\text{Dom}_f = (-\infty, 4]$ and the range of $f$ is $\text{Ran}_f = [0, \infty)$.

However, there is an obvious danger here. If you do not keep the restriction that $y = \sqrt{4 - x} \geq 0$ in mind, you could come to the false conclusion that the domain of $g(x) = 4 - x^2$ is $(-\infty, \infty)$ rather than $[0, \infty)$.